3 Phase Current and Voltage Drop 1

[djr3203]

I am a bit rusty with 3 phase calculations and I was hoping I could get some input.

In calculating current for 3 phase the equation is I=VA/(V*1.73) [not taking into account PF]. If you have a 3 phase wye distribution, does this then give you the current traveling through each phase-to-neutral?

When calculating voltage drop in conductors smaller than 1/0, the equation is Vd=(1.732xKxIxD)/CM. I have 2000KVA transformer with a 480V 4 wire wye secondary, and the documents call for 8 conduits with 4#600kcmil wires would I take 2000KVA/(1.73x480) and up it 15% since the transformers are outdoor padmount to get approx. 2700A. Would I then divide that current by 8 to get approx 337A for each wire (assuming all the wires are the same length)? If I then used that in the above voltage drop calculation, would that be the voltage drop on each wire for each phase?

Please be gentle i'm a new, junior engineer

David

 

[djr3203]

I apologize. I meant 480V delta secondary on the transformer, not wye.

 

[djr3203]

Good gosh it is wye... I was right the first time.

 

[rcw retired EE]

You are on the right track and will get answers in the ball park, ignoring things like power factor, unequal current distribution and variations in cable impedance.

With larger cable sizes the cable reactive impedance has more of an effect on voltage drop than the resistance. The Vdrop equation you have will not be as accurate. Look on line for some tables of cable impedance and/or voltage drop. Okonite or Southworth are a good place to start. Also check Eaton's Consulting Engineering Handbook.

Usually, we don't worry too much about the voltage drop of the transformer secondary cables. It is small compared to the drop in the rest of the system.

http://www.eaton.com/Electrical/Consultants/ConsultingApplicationGuide/index.htm

 

[MAS2000]

I = approx. 2700A per phase is ok 480V Y secondary

rcwilson is right that volt drop is dependant on both the resistance & reactance; so you should use the cable impedance per foot or per metre instead of just using resistance in volt drop calculations.

To elaborate more you should explain where and for what purposes the volt-drop is required? Do you want to calculate volt-drop when transformer is fully loaded (with 15% overload)? What are these conduits feeding? Are they feeding a distribution board or the load? More information is required. A drawing will be appreciated.

The voltage drop will depend on the load at end of the cable system. For balanced load conditions (and also assuming that all cables are similar in type and length); the volt drop per phase will be almost similar; On other hand it won't be similar for unbalanced load conditions and calculations would be more complex.

If your system feeds single phase loads; the volt drop per phase will be twice for the single phase circuit; this accounts for the drop in phase and return conductor.

M.A.Sh.
Elect. Engr.

 

[djr3203]

Thanks for the answer. Just to confirm, the calculated 2700A is then for each phase as mas2000 said? How does waross's calculation of 7220A per phase come in to play?

Waross, are you sure your calcs are on point with the current? The wire gauge was taken out of our standard feeder schedule which has been used for a long time without any problems or anyone else bringing up a red flag.

The padmount transformers are located quite some distance away from the building and will be the service feed for the building. They will feed a switchgear in the basement. The client was beginning to worry about voltage drop since the transformers were located so far away (approx. 250' away from the building).

I was assuming to do the calculations at sort of a "worst case" scenario with the transformers being overloaded by 15%.

Once we circuit, we will do it as such to balance the loads on all the phases.

 

[MAS2000]

djr3203; can your customer's load be assumed balanced 3-phase?

Are there 8 cables per phase or 2 cables per phase + 2 cables for neutral between the transformer & the customer's intake?

M.A.Sh.
Elect. Engr.

 

[djr3203]

The building is new construction, so I would say yes the load can be assumed to be balanced.

Between the transformer and the switchgear there will be 8 cables per phase. Currently the documents state (8)4"C-(4)#600KCMIL.

 

[waross]

Should be 2407 amps.
Divide by 8 = 301 Amps per phase. (Plus 15% = 346 Amps)
My tables show 600 MCM as allowable for:
60 Deg. C 355 Amps
75 Deg. C 420 Amps
90 Deg. C 455 Amps
Cable sizes ampacities are based on temperature.
If the diameter of a length of conductor is doubled, the cross sectional area is quadrupled but the surface area is doubled.
As a result, allowable current densities drop of as cable diameter increases. When the currents get up into hundreds of amps, skin effect causes a further reduction in the allowable current average density. (skin effect works with I²R_effective to add heat.)
The effect of this is that Voltage drop is seldom an issue with large feeders and moderate lengths.
I find it easier to understand if I use the line to neutral voltage to calculate the voltage drop of one phase to neutral. Use the phase current and subtract the voltage drop from 277 Volts. (Just calculate the phase conductor voltage drop, ignore the neutral.)
Now you may calculate the percentage drop which will also be valid for the phase to phase drop, or you may multiply the single phase drop in Volts by [√]3 to get the line to line voltage drop.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[jghrist]

rcwilson is right that volt drop is dependant on both the resistance & reactance; so you should use the cable impedance per foot or per metre instead of just using resistance in volt drop calculations.

Where cable reactance is significant and/or the power factor is significantly different than 100%, the equation is not correct. You should use VD = I·[R·cos(theta) + X·sin(theta)] where cos(theta) = power factor.

 

[7anoter4]

First of all in order to keep the same current intensity in each single core cable you have to keep the phase order
- a certain order. That means-if there are 4 layers of 2 cables per phase, for instance, you have to lay them so:
A B C C B A in each layer. Changing the order one or more cables will be overloaded and other under loaded.
Second, the conduit has to be PVC made [not steel] and no rebar shall be close loop around the conduit.
The duct bank is buried in soil and the soil thermal resistivity may be different from 90 w/cm/oC as per NEC, for instance.
But NEC is limited to 4 single core cables per phase so it is not a good source of information.
You may calculate the ampacity following Neher theory or IEC60287.
So, if the soil and concrete thermal resistance will be 90 w/cm/oC ,the distance between ducts [conduits] will be 10 inch at least, the surrounded soil temperature will be 20 oC and the duct bank top depth will be not more than 2 ft then at 345.8 A per each cable the conductor temperature will be not more than 90 oC.[for 600 MCM copper 4 rows 6 columns].
Voltage drop up to transformer terminals is already 5.1% [for 6% short-circuit impedance and approx. 20 kw copper losses].
Let's say you need on switchgear busbar 12.8 kV in the full load case.
The cable resistance will be as waross explained, but the reactance is more complicate.
If we would take the reactance for a single group in flat arrangement for 250 ft length we could get the following:
Rcabl= 0.006108 ohm Xcabl= 0.019881 ohm.
With these data we would get 2.1% voltage drop on cable and total drop could be 7.22%.

 

[waross]

Some comments:
With four 600 MCM cables per conduit I am assuming three phase conductors and a neutral in each conduit. A little overkill on the neutral but completely acceptable. Magnetic encirclement of each conduit is not an issue.
So steel conduit is perfectly acceptable.

PU impedance describes current under short circuit conditions and X predominates.
At full load current transformer regulation describes voltage drop and R predominates and is much less than X or Z. for unity power factor loads. At 90% PF, the transformer voltage drop is often greater but still less than a calculation based on Z.

In the case of three cables per phase in close proximity the reactance is less than with spaced cables. The cables touching and in trefoil arrangement is best but that may not be assumed when cables are pulled into conduit especially with a fourth, neutral cable.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[7anoter4]

You are right waross !
It is possible to pull 4* 600 MCM cables[insulated and jacketed] in a 4" steel conduit as the filling factor is only 34%-
less than 40% maximum permitted by NEC ch.9.
Now, 8 parallel conduits with 3*600 mcm copper in a 90 w/cm/oC soil at 2 ft depth will support 345.8 A per cable without reach 90oC. But will be in a single row 10 inch apart one from another.
That will solve the voltage drop. In previous post I forgot to divide by 8 so the voltage drop across the cable was anyway 2.1/8=0.2625%.Now will be less.
I forgot it is not a power station so one does not need to be so careful. We don't use more than 30% filling pulling the cable through a conduit and I never take less than 120 w/cm/oC soil resistivity.
Briefly, your indication it is a very good solution.

 

[waross]

Thank you for the kind words 7anoter4. In a previous post I forgot to divide by three phases. It happens.
Yours
Bill

Bill
--------------------
"Why not the best?"
Jimmy Carter